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The general iterative methods for equilibrium problems and fixed point problems of a countable family of nonexpansive mappings in Hilbert spaces

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Abstract

In this paper, the researcher introduces the general iterative scheme for finding a common element of the set of equilibrium problems and fixed point problems of a countable family of nonexpansive mappings in Hilbert spaces. The results presented in this paper improve and extend the corresponding results announced by many others.

MSC:47H10, 47H09.

1 Introduction

Let H be a real Hilbert space with the inner product , and the norm . Let C be a nonempty closed and convex subset of H, and let T:CC be a nonlinear mapping. In this paper, we use F(T) to denote the fixed point set of T.

Recall the following definitions.

  1. (1)

    The mapping T is said to be nonexpansive if

    TxTyxy,x,yC.
    (1.1)

Further, let F be a bifunction from C×C into , where is the set of real numbers. The so-called equilibrium problem for F:C×CR is to find yC such that

F(y,u)0,uC.
(1.2)

The set of solutions of (1.2) is denoted by EP(F). Given a mapping A:CH, let F(y,u)=Ay,uy for all y,uC. Then zEP(F) if and only if Az,uz0 for all uC. Numerous problems in physics, optimization and economics reduce to finding a solution of (1.2).

  1. (2)

    The mappings { T n } n N are said to be a family of nonexpansive mappings from H into itself if

    T n x T n yxy,x,yH,
    (1.3)

and denoted by F( T n )={xH: T n x=x} is the fixed point set of T n . Finding an optimal point in n N F( T n ) of the fixed point sets of each mapping is a matter of interest in various branches of science.

Recently, many authors considered the iterative methods for finding a common element of the set of solutions to problem (1.2) and of the set of fixed points of nonexpansive mappings; see, for example, [1, 2] and the references therein.

Next, let A:CH be a nonlinear mapping. We recall the following definitions.

  1. (3)

    A is said to be monotone if

    AxAy,xy0,x,yC.
  2. (4)

    A is said to be strongly monotone if there exists a constant α>0 such that

    AxAy,xyα x y 2 ,x,yC.

In such a case, A is said to be α-strongly monotone.

  1. (5)

    A is said to be inverse-strongly monotone if there exists a constant α>0 such that

    AxAy,xyα A x A y 2 ,x,yC.

In such a case, A is said to be α-inverse-strongly monotone.

The classical variational inequality is to find uC such that

Au,vu0,vC.
(1.4)

In this paper, we use VI(C,A) to denote the set of solutions to problem (1.4). One can easily see that the variational inequality problem is equivalent to a fixed point problem. uC is a solution to problem (1.4) if and only if u is a fixed point of the mapping P C (Iλ)T, where λ>0 is a constant.

The variational inequality has been widely studied in the literature; see, for example, the work of Plubtieng and Punpaeng [3] and the references therein.

Recently, Ceng et al. [4] considered an iterative method for the system of variational inequalities (1.4). They got a strongly convergence theorem for problem (1.4) and a fixed point problem for a single nonexpansive mapping; see [4] for more details.

On the other hand, Moudafi [5] introduced the viscosity approximation method for nonexpansive mappings (see [6] for further developments in both Hilbert and Banach spaces).

A mapping f:CC is called α-contractive if there exists a constant α(0,1) such that

f ( x ) f ( y ) αxy,x,yC.
(1.5)

Let f be a contraction on C. Starting with an arbitrary initial x 1 C, define a sequence { x n } recursively by

x n + 1 =(1 σ n )T x n + σ n f( x n ),n0,
(1.6)

where { σ n } is a sequence in (0,1). It is proved [5, 6] that under certain appropriate conditions imposed on { σ n }, the sequence { x n } generated by (1.6) strongly converges to the unique solution q in C of the variational inequality

( I f ) q , p q 0,pC.

Let A be a strongly positive linear bounded operator on a Hilbert space H with a constant  γ ¯ ; that is, there exists γ ¯ >0 such that

Ax,x γ ¯ x 2 ,xH.
(1.7)

Recently, Marino and Xu [7] introduced the following general iterative method:

x n + 1 =(I α n A)T x n + α n γf( x n ),n0,
(1.8)

where A is a strongly positive bounded linear operator on H. They proved that if the sequence { α n } of parameters satisfies appropriate conditions, then the sequence { x n } generated by (1.8) converges strongly to the unique solution of the variational inequality

( A γ f ) x , x x 0,xC,
(1.9)

which is the optimality condition for the minimization problem

min x C 1 2 Ax,xh(x),

where h is a potential function for γf (i.e., h (x)=γf(x) for xH).

In 2007, Takahashi and Takahashi [2] introduced an iterative scheme by the viscosity approximation method for finding a common element of the set of solutions (1.2) and the set of fixed points of a nonexpansive mapping in Hilbert spaces. Let S:CH be a nonexpansive mapping. Starting with arbitrary initial x 1 H, define sequences { x n } and { u n } recursively by

{ F ( u n , y ) + 1 r n y u n , u n x n 0 , y C , x n + 1 = α n γ f ( x n ) + ( 1 α n ) S u n , n N .
(1.10)

They proved that under certain appropriate conditions imposed on { α n } and { r n }, the sequences { x n } and { u n } converge strongly to zF(S)EP(F), where z= P F ( S ) E P ( F ) f(z).

Next, Plubtieng and Punpaeng, [3] introduced an iterative scheme by the general iterative method for finding a common element of the set of solutions (1.2) and the set of fixed points of nonexpansive mappings in Hilbert spaces.

Let S:HH be a nonexpansive mapping. Starting with an arbitrary x 1 H, define sequences { x n } and { u n } by

{ F ( u n , y ) + 1 r n y u n , u n x n 0 , y C , x n + 1 = α n γ f ( x n ) + ( I α n A ) S u n , n N .
(1.11)

They proved that if the sequences { α n } and { r n } of parameters satisfy appropriate conditions, then the sequence { x n } generated by (1.11) converges strongly to the unique solution of the variational inequality

( A γ f ) z , x z 0,xF(S)EP(F),
(1.12)

which is the optimality condition for the minimization problem

min x F ( S ) E P ( F ) 1 2 Ax,xh(x),

where h is a potential function for γf (i.e., h (x)=γf(x) for xH).

Let T 1 , T 2 , be an infinite sequence of mappings of C into itself, and let λ 1 , λ 2 , be real numbers such that 0 λ i 1 for every iN. Then for any nN, Takahashi [8] (see [9]) defined a mapping W n of C into itself as follows:

(1.13)

Such a mapping W n is called the W-mapping generated by T n , T n 1 ,, T 1 and λ n , λ n 1 ,, λ 1 .

Recently, using process (1.13), Yao et al. [10] proved the following result.

Theorem 1.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let F:C×CR be an equilibrium bifunction satisfying the conditions:

  1. (1)

    F is monotone, that is, F(x,y)+F(y,x)0 for all x,yC;

  2. (2)

    for each x,y,zC, lim t 0 F(tz+(1t)x,y)F(x,y);

  3. (3)

    for each xC, yF(x,y) is convex and lower semicontinuous.

Let { T i } i = 1 be an infinite family of nonexpansive mappings of C into C such that i = 1 F( T i )EP(F). Suppose { α n }, { β n } and { γ n } are three sequences in (0,1) such that α n + β n + γ n =1 and { r n }(0,). Suppose the following conditions are satisfied:

  1. (1)
    lim n α n =0

    and n = 1 α n =;

  2. (2)
    0< lim inf n β n lim sup n β n <1

    ;

  3. (3)
    lim inf n r n >0

    and lim n ( r n + 1 r n )=0.

Let f be a contraction of H into itself, and let x 0 H be given arbitrarily. Then the sequences { x n } and { y n } generated iteratively by

{ F ( y n , x ) + 1 r n x y n , y n x n 0 , x C , x n + 1 = α n f ( x n ) + β n x n + γ n W n y n ,

converge strongly to x i = 1 F( T i )EP(F), the unique solution of the minimization problem

min x i = 1 F ( T i ) E P ( F ) 1 2 x 2 h(x),

where h is a potential function for f.

Very recently, using process (1.13), Chen [11] proved the following result.

Theorem 1.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let { T n } n = 1 be a sequence of nonexpansive mappings from C to C such that the common fixed point set Ω= n = 1 F( T n ). Let f:CH be an α-contraction, and let A:HH be a self-adjoint, strongly positive bounded linear operator with a coefficient γ ¯ >0. Let σ be a constant such that 0<γα< γ ¯ . For an arbitrary initial point x 0 belonging to C, one defines a sequence { x n } n 0 iteratively

x n + 1 = P C [ α n γ f ( x n ) + ( I α n A ) W n x n ] ,n0,
(1.14)

where { α n } is a real sequence in [0,1]. Assume the sequence { α n } satisfies the following conditions:

(C1) lim n α n =0;

(C2) n = 1 α n =.

Then the sequence { x n } generated by (1.14) converges in norm to the unique solution x , which solves the following variational inequality:

x Ω such that ( A γ f ) x , x x ˆ 0, x ˆ Ω.
(1.15)

Motivated by this result, we introduce the following explicit general iterative scheme:

{ x 1 H , F ( u n , y ) + 1 r n y u n , u n x n 0 , y H , x n + 1 = P C [ α n γ f ( x n ) + ( I α n A ) W n u n ] , n N ,
(1.16)

where { T n } n N is a family of nonexpansive mappings from H into itself such that n N F( T n ) is nonempty, F:C×CR is an equilibrium bifunction, A is a strongly positive operator on H, f is a contraction of H into itself with α(0,1), { α n }, { r n }, { λ n } suitable sequences in and { W n } is the sequence of a W-mapping generated by { T n } n N and { λ n }. Let U be defined by Ux= lim n W n x= lim n U n , 1 x for every xC using process (1.13). We shall prove under mild conditions that { x n } and { u n } strongly converge to a point x i = 1 F( T i )EP(F), which is the unique solution of the variational inequality

( A γ f ) x , x x ˆ 0, x ˆ i = 1 F( T i )EP(F),
(1.17)

or, equivalently, the unique solution of the minimization problem

min x i = 1 F ( T i ) E P ( F ) 1 2 A x ˆ , x ˆ h( x ˆ ),

where h is a potential function for γf.

2 Preliminaries

Let H be a real Hilbert space with the norm and the inner product ,, and let C be a closed convex subset of H. We call f:CH an α-contraction if there exists a constant α[0,1) such that

f ( x ) f ( y ) αxy,x,yC.

Let A be a strongly positive linear bounded operator on a Hilbert space H with a constant  γ ¯ ; that is, there exists γ ¯ >0 such that

Ax,x γ ¯ x 2 ,xH.

Next, we denote weak convergence and strong convergence by notations and →, respectively. A space X is said to satisfy Opial’s condition [12] if for each sequence { x n } in X which converges weakly to a point xX, we have

lim inf n x n x< lim inf n x n y,yX,yx.

For every point xH, there exists a unique nearest point in C, denoted by P C x, such that

x P C xxy,yC.
P C

is called the (nearest point or metric) projection of H onto C. In addition, P C x is characterized by the following properties: P C xC and

(2.1)
(2.2)

Recall that a mapping T:HH is said to be firmly nonexpansive mapping if

T x T y 2 TxTy,xy,x,yH.

It is well known that P C is a firmly nonexpansive mapping of H onto C and satisfies

P C x P C y 2 xy, P C x P C y,x,yH.
(2.3)

If A is an α-inverse-strongly monotone mapping of C into H, then it is obvious that A is 1 α -Lipschitz continuous. We also have that for all x,yC and λ>0,

( I λ A ) x ( I λ A ) y 2 = x y λ ( A x A y ) 2 = x y 2 2 λ A x A y , x y + λ 2 A x A y 2 x y 2 + λ ( λ 2 α ) A x A y 2 .
(2.4)

So, if λ2α, then IλA is a nonexpansive mapping of C into H.

The following lemmas will be useful for proving the convergence result of this paper.

Lemma 2.1 Let H be a real Hilbert space. Then for all x,yH,

  1. (1)
    x + y 2 x 2 +2y,x+y

    ;

  2. (2)
    x + y 2 x 2 +2y,x

    .

Lemma 2.2 ([13])

Let { x n } and { y n } be bounded sequences in a Banach space X, and let { β n } be a sequence in [0,1] with 0< lim inf n β n lim sup n β n <1. Suppose that x n + 1 =(1 β n ) y n + β n x n for all integers n0 and lim sup n ( y n + 1 y n x n + 1 x n )0. Then lim n y n x n =0.

Lemma 2.3 ([14])

Assume that F:C×CR, let us assume that F satisfies the following conditions:

(A1) F(x,x)=0 for all xC;

(A2) F is monotone, i.e., F(x,y)+F(y,x)0 for all x,yC;

(A3) for each x,y,zC, lim t 0 F(tz+(1t)x,y)F(x,y);

(A4) for each xC, yF(x,y) is convex and lower semicontinuous.

Lemma 2.4 ([14])

Assume that F:C×CR satisfies (A1)-(A4). For r>0 and xH, define a mapping T r :HC as follows:

T r (x)= { z C : F ( z , y ) + 1 r y z , z x 0 , y C }

for all zH. Then the following hold:

  1. 1.
    T r

    is single-valued;

  2. 2.
    T r

    is firmly nonexpansive, i.e., for any x,yH,

    T r x T r y 2 T r x T r y,xy;
  3. 3.
    F( T r )=EP(F)

    ;

  4. 4.

    EP(F) is closed and convex.

Lemma 2.5 ([12])

Let H be a Hilbert space, C be a closed convex subset of H, and S:CC be a nonexpansive mapping with F(S). If { x n } is a sequence in C weakly converging to xC and if {(IS) x n } converges strongly to y, then (IS)x=y.

Lemma 2.6 ([6])

Assume that { a n } is a sequence of nonnegative real numbers such that

a n + 1 (1 γ n ) a n + δ n ,n0,

where { γ n } is a sequence in (0,1) and { δ n } is a sequence in such that

  1. (1)
    n = 1 γ n =

    ;

  2. (2)
    lim sup n δ n γ n 0

    or n = 1 | δ n |<.

Then lim n a n =0.

Lemma 2.7 ([7])

Let H be a Hilbert space, C be a nonempty closed convex subset of H, and f:HH be a contraction with a coefficient 0<α<1, and let A be a strongly positive linear bounded operator with a coefficient γ ¯ >0. Then, for 0<γ< γ ¯ α ,

x y , ( A γ f ) x ( A γ f ) y ( γ ¯ γα) x y 2 ,x,yH.

That is, Aγf is strongly monotone with a coefficient γ ¯ γα.

Lemma 2.8 ([7])

Assume A is a strongly positive linear bounded operator on a Hilbert space H with a coefficient γ ¯ >0 and 0<ρ A 1 . Then IρA1ρ γ ¯ .

Lemma 2.9 ([9] and [15])

Let C be a nonempty closed convex subset of a Banach space E. Let { T i } i = 1 be a sequence of nonexpansive mappings of C into itself with i = 1 F( T i ), and let { λ i } i = 1 be a real sequence such that 0< λ i b<1, i1. Then:

  1. (1)
    W n

    is nonexpansive and F( W n )= i = 1 F( T i ) for each n1;

  2. (2)

    for each xC and for each positive integer k, the lim n U n , k x exists;

  3. (3)

    the mapping U:CC defined by

    Ux= lim n W n x= lim n U n , 1 x,xC

is a nonexpansive mapping satisfying F(U)= n = 1 F( T i ) and it is called the W-mapping generated by T 1 , T 2 , and λ 1 , λ 2 , ;

  1. (4)
    lim m , n sup x K W m x W n x=0

    for any bounded subset K of E.

3 Main results

In this section, we introduce our algorithm and prove its strong convergence.

Theorem 3.1 Let C be a closed convex subset of a real Hilbert space H. Let F be a bifunction from H×H into satisfying (A1)-(A4). Let f be a contraction of H into itself with α(0,1), and let T n be a sequence of nonexpansive mappings of C into itself such that Ω= n = 1 F( T n )EP(F). Let A:HH be a strongly positive bounded linear operator with a coefficient γ ¯ >0 with 0<γ< γ ¯ α . Let λ 1 , λ 2 , be a sequence of real numbers such that 0< λ n b<1 for every n=1,2, . Let W n be a W-mapping of C into itself generated by T n , T n 1 ,, T 1 and λ n , λ n 1 ,, λ 1 . Let U be defined by Ux= lim n W n x= lim n U n , 1 x for every xC. Let { x n } and { u n } be sequences generated by x 1 H and

{ F ( u n , y ) + 1 r n y u n , u n x n 0 , y H , x n + 1 = P C [ α n γ f ( x n ) + ( I α n A ) W n u n ] , n N ,
(3.1)

where { α n } is a sequence in (0,1) and { r n } is a sequence in [0,). Suppose that { α n } and { r n } satisfy the following conditions:

(C1) lim n α n =0;

(C2) n = 1 α n =;

(C3) lim n r n =r>0.

Then both { x n } and { u n } converge strongly to x Ω, which is the unique solution of the variational inequality

( A γ f ) x , x x ˆ 0, x ˆ Ω.
(3.2)

Equivalently, one has x = P Ω (IA+γf)( x ).

Proof We observe that P Ω (γf+(IA)) is a contraction. Indeed, for all x,yH, we have

P Ω ( γ f + ( I A ) ) ( x ) P Ω ( γ f + ( I A ) ) ( y ) ( γ f + ( I A ) ) ( x ) ( γ f + ( I A ) ) ( y ) γ f ( x ) f ( y ) + I A x y γ α x y + ( 1 γ ¯ ) x y < ( 1 ( γ ¯ γ α ) ) x y .

Banach’s contraction mapping principle guarantees that P Ω (γf+(IA)) has a unique fixed point, say x H. That is, x = P Ω (γf+(IA))( x ). Note that by Lemma 2.4, we can write

x n + 1 = P C [ α n γ f ( x n ) + ( I α n A ) W n T r n x n ] ,

where

T r n (x)= { z H : F ( z , y ) + 1 r n y z , z x 0 , y H } .

Moreover, since α n 0 as n by condition (C1), we assume that α n A 1 for all nN. From Lemma 2.8, we know that if 0<ρ< A 1 , then IρA1ρ γ ¯ . We divide the proof into seven steps as follows.

Step 1. Show that the sequences { x n } and { u n } are bounded.

Let x ˆ Ω. Then x ˆ EP(F). From Lemma 2.4, we have

u n x ˆ = T r n x n T r n x ˆ x n x ˆ .

Thus, we have

x n + 1 x ˆ = P C [ α n γ f ( x n ) + ( I α n A ) W n u n ] x ˆ α n γ f ( x n ) + ( I α n A ) W n u n x ˆ α n γ f ( x n ) f ( x ˆ ) + I α n A W n u n x ˆ + α n γ f ( x ˆ ) A x ˆ α n γ α x n x ˆ + ( 1 α n γ ¯ ) u n x ˆ + α n γ f ( x ˆ ) A x ˆ = ( 1 α n ( γ ¯ γ α ) ) x n x ˆ + α n ( γ ¯ γ α ) γ f ( x ˆ ) A x ˆ ( γ ¯ γ α ) .

By induction, we have

x n x ˆ max { x 1 x ˆ , γ f ( x ˆ ) A x ˆ γ ¯ γ α } ,n0.

This shows that the sequence { x n } is bounded, so are { u n }, {f( x n )} and { W n u n }.

Step 2. Show that W n + 1 u n W n u n 0 as n.

Let x ˆ Ω. Since T i and U n , i are nonexpansive and T i x ˆ = x ˆ = U n , i x ˆ for every nN and in+1, it follows that

W n + 1 u n W n u n = λ 1 T 1 U n + 1 , 2 u n λ 1 T 1 U n , 2 u n λ 1 U n + 1 , 2 u n U n , 2 u n = λ 1 λ 2 T 2 U n + 1 , 3 u n λ 2 T 2 U n , 3 u n λ 1 λ 2 U n + 1 , 3 u n U n , 3 u n ( i = 1 n λ i ) U n + 1 , n + 1 u n x ˆ + x ˆ U n , n + 1 u n ( i = 1 n λ i ) ( U n + 1 , n + 1 u n x ˆ + x ˆ U n , n + 1 u n ) 2 ( i = 1 n λ i ) u n x ˆ .

Since { u n } is bounded and 0< λ n b<1 for any nN, the following holds:

lim n W n + 1 u n W n u n =0.

Step 3. Show that x n + 1 x n 0 as n.

Setting S=2 P C I, we have S is nonexpansive. Note that W n =(1 λ 1 )I+ λ 1 T 1 U n , 2 . Then we can write

x n + 1 = I + S 2 [ α n γ f ( x n ) + ( I α n A ) W n u n ] = 1 α n 2 W n u n + α n 2 ( γ f ( x n ) A W n u n + W n u n ) + 1 2 S [ α n γ f ( x n ) + ( I α n A ) W n u n ] = 1 α n 2 [ ( 1 λ 1 ) I + λ 1 T 1 U n , 2 ] u n + α n 2 ( γ f ( x n ) A W n u n + W n u n ) + 1 2 S [ α n γ f ( x n ) + ( I α n A ) W n u n ] = ( 1 λ 1 ) ( 1 α n ) 2 u n + λ 1 ( 1 α n ) 2 T 1 U n , 2 u n + α n 2 ( γ f ( x n ) A W n u n + W n u n ) + 1 2 S [ α n γ f ( x n ) + ( I α n A ) W n u n ] .
(3.3)

Note that

0< lim n ( 1 λ 1 ) ( 1 α n ) 2 = 1 λ 1 2 <1,

and

λ 1 ( 1 α n ) 2 + 1 2 = 1 + λ 1 2 λ 1 2 α n .

From (3.3), we have

x n + 1 = [ 1 ( 1 + λ 1 2 + 1 λ 1 2 α n ) ] u n + ( 1 + λ 1 2 + 1 λ 1 2 α n ) × ( λ 1 ( 1 α n ) 2 T 1 U n , 2 u n + α n 2 ( γ f ( x n ) A W n u n + W n u n ) + 1 2 S [ α n γ f ( x n ) + ( I α n A ) W n u n ] ) / ( 1 + λ 1 2 + 1 λ 1 2 α n ) = [ 1 ( 1 + λ 1 2 + 1 λ 1 2 α n ) ] u n + ( 1 + λ 1 2 + 1 λ 1 2 α n ) y n ,
(3.4)

where

y n = ( λ 1 ( 1 α n ) 2 T 1 U n , 2 u n + α n 2 ( γ f ( x n ) A W n u n + W n u n ) + 1 2 S [ α n γ f ( x n ) + ( I α n A ) W n u n ] ) / ( 1 + λ 1 2 + 1 λ 1 2 α n ) = ( λ 1 ( 1 α n ) T 1 U n , 2 u n + α n ( γ f ( x n ) A W n u n + W n u n ) + S [ α n γ f ( x n ) + ( I α n A ) W n u n ] ) / ( 1 + λ 1 + ( 1 λ 1 ) α n ) .

Set e n =γf( x n )A W n u n + W n u n and d n = α n γf( x n )+(I α n A) W n u n for all n. Then

y n = λ 1 ( 1 α n ) T 1 U n , 2 u n + α n e n + S d n 1 + λ 1 + ( 1 λ 1 ) α n ,n0.

It follows that

y n + 1 y n = λ 1 ( 1 α n + 1 ) T 1 U n + 1 , 2 u n + 1 + α n + 1 e n + 1 + S d n + 1 1 + λ 1 + ( 1 λ 1 ) α n + 1 λ 1 ( 1 α n ) T 1 U n , 2 u n + α n e n + S d n 1 + λ 1 + ( 1 λ 1 ) α n = λ 1 ( 1 α n + 1 ) 1 + λ 1 + ( 1 λ 1 ) α n + 1 ( T 1 U n + 1 , 2 u n + 1 T 1 U n , 2 u n ) + ( λ 1 ( 1 α n + 1 ) 1 + λ 1 + ( 1 λ 1 ) α n + 1 λ 1 ( 1 α n ) 1 + λ 1 + ( 1 λ 1 ) α n ) T 1 U n , 2 u n + α n + 1 e n + 1 1 + λ 1 + ( 1 λ 1 ) α n + 1 α n e n 1 + λ 1 + ( 1 λ 1 ) α n + S d n + 1 S d n 1 + λ 1 + ( 1 λ 1 ) α n + 1 + ( 1 1 + λ 1 + ( 1 λ 1 ) α n + 1 1 1 + λ 1 + ( 1 λ 1 ) α n ) S d n .

Thus,

y n + 1 y n λ 1 ( 1 α n + 1 ) 1 + λ 1 + ( 1 λ 1 ) α n + 1 T 1 U n + 1 , 2 u n + 1 T 1 U n , 2 u n + | λ 1 ( 1 α n + 1 ) 1 + λ 1 + ( 1 λ 1 ) α n + 1 λ 1 ( 1 α n ) 1 + λ 1 + ( 1 λ 1 ) α n | T 1 U n , 2 u n + α n + 1 1 + λ 1 + ( 1 λ 1 ) α n + 1 e n + 1 + α n 1 + λ 1 + ( 1 λ 1 ) α n e n + 1 1 + λ 1 + ( 1 λ 1 ) α n + 1 S d n + 1 S d n + | 1 1 + λ 1 + ( 1 λ 1 ) α n + 1 1 1 + λ 1 + ( 1 λ 1 ) α n | S d n .

Since S is nonexpansive, we obtain that

S d n + 1 S d n d n + 1 d n = α n + 1 γ f ( x n + 1 ) + ( I α n + 1 A ) W n + 1 u n + 1 ( α n γ f ( x n ) + ( I α n A ) W n u n ) α n + 1 γ f ( x n + 1 ) A W n + 1 u n + 1 + α n γ f ( x n ) A W n u n ) + W n + 1 u n + 1 W n u n α n + 1 γ f ( x n + 1 ) A W n + 1 u n + 1 + α n γ f ( x n ) A W n u n ) + W n + 1 u n + 1 W n + 1 u n + W n + 1 u n W n u n α n + 1 γ f ( x n + 1 ) A W n + 1 u n + 1 + α n γ f ( x n ) A W n u n ) + u n + 1 u n . + W n + 1 u n W n u n .

Since T i and U n , i are nonexpansive, we have

T 1 U n + 1 , 2 u n T 1 U n , 2 u n U n + 1 , 2 u n U n , 2 u n = λ 2 T 2 U n + 1 , 3 u n λ 2 T 2 U n , 3 u n λ 2 U n + 1 , 3 u n U n , 3 u n λ 2 λ n U n + 1 , n + 1 u n U n , n + 1 u n M i = 2 n λ i ,

where M>0 is a constant such that U n + 1 , n + 1 u n U n , n + 1 u n M for all n0. So,

T 1 U n + 1 , 2 u n + 1 T 1 U n , 2 u n T 1 U n + 1 , 2 u n + 1 T 1 U n + 1 , 2 u n + T 1 U n + 1 , 2 u n T 1 U n , 2 u n u n + 1 u n + M i = 2 n λ i .

Hence,

y n + 1 y n λ 1 ( 1 α n + 1 ) 1 + λ 1 + ( 1 λ 1 ) α n + 1 u n + 1 u n + M i = 2 n λ i + | λ 1 ( 1 α n + 1 ) 1 + λ 1 + ( 1 λ 1 ) α n + 1 λ 1 ( 1 α n ) 1 + λ 1 + ( 1 λ 1 ) α n | T 1 U n , 2 u n + α n + 1 1 + λ 1 + ( 1 λ 1 ) α n + 1 e n + 1 + α n 1 + λ 1 + ( 1 λ 1 ) α n e n + 1 1 + λ 1 + ( 1 λ 1 ) α n + 1 ( α n + 1 γ f ( x n + 1 ) A W n + 1 u n + 1 + α n γ f ( x n ) A W n u n + u n + 1 u n ) + W n + 1 u n W n u n + | 1 1 + λ 1 + ( 1 λ 1 ) α n + 1 1 1 + λ 1 + ( 1 λ 1 ) α n | S d n = λ 1 ( 1 α n + 1 ) 1 + λ 1 + ( 1 λ 1 ) α n + 1 u n + 1 u n + | λ 1 ( 1 α n + 1 ) 1 + λ 1 + ( 1 λ 1 ) α n + 1 λ 1 ( 1 α n ) 1 + λ 1 + ( 1 λ 1 ) α n | T 1 U n , 2 u n + α n + 1 1 + λ 1 + ( 1 λ 1 ) α n + 1 e n + 1 + α n 1 + λ 1 + ( 1 λ 1 ) α n e n + 1 1 + λ 1 + ( 1 λ 1 ) α n + 1 ( α n + 1 γ f ( x n + 1 ) A W n + 1 u n + 1 + α n γ f ( x n ) A W n u n + u n + 1 u n ) + W n + 1 u n W n u n + | 1 1 + λ 1 + ( 1 λ 1 ) α n + 1 1 1 + λ 1 + ( 1 λ 1 ) α n | S d n .

Note that: (1)

By condition (C1), we have

λ 1 ( 1 α n + 1 ) 1 + λ 1 + ( 1 λ 1 ) α n + 1 λ 1 ( 1 α n ) 1 + λ 1 + ( 1 λ 1 ) α n 0

and

1 1 + λ 1 + ( 1 λ 1 ) α n + 1 1 1 + λ 1 + ( 1 λ 1 ) α n 0.
  1. (2)
    W n + 1 u n W n u n 0

    as n because of Step 2.

Therefore,

lim sup n ( y n + 1 y n u n + 1 u n ) 0.

By Lemma 2.2, we get

lim n y n u n =0.

Hence, from (3.4), we deduce

lim n x n + 1 x n = lim n ( 1 + λ 1 2 + 1 λ 1 2 α n ) y n u n =0.
(3.5)

Step 4. Show that x n W n u n 0 as n. Indeed, we have

x n W n u n x n x n + 1 + x n + 1 W n u n = x n x n + 1 + P C [ α n γ f ( x n ) + ( I α n A ) W n u n ] W n u n x n x n + 1 + α n γ f ( x n ) + ( I α n A ) W n u n W n u n = x n x n + 1 + α n A W n u n + α n γ f ( x n ) x n x n + 1 + α n ( A W n u n + γ f ( x n ) ) .

Then

lim n x n W n u n lim n x n x n + 1 + α n ( A W n u n + γ f ( x n ) ) =0.
(3.6)

Thus, from (3.6), we obtain

lim n x n W n u n =0.

Step 5. Show that x n u n 0 as n.

Let x ˆ Ω. Since T r n is firmly nonexpansive, it follows

x ˆ u n 2 = T r n x ˆ T r n x n 2 T r n x n T r n x ˆ , x n x ˆ T r n x n x ˆ , x n x ˆ = u n x ˆ , x n x ˆ = 1 2 ( u n x ˆ 2 + x n x ˆ 2 x n u n 2 ) .

Then

u n x ˆ 2 x n x ˆ 2 x n u n 2 .

Since we have

x n + 1 x ˆ 2 = P C [ α n γ f ( x n ) + ( I α n A ) W n u n ] P C x ˆ 2 = α n γ f ( x n ) + ( I α n A ) W n u n x ˆ 2 = ( I α n A ) ( W n u n x ˆ ) + α n ( γ f ( x n ) A x ˆ ) 2 ( 1 α n γ ¯ ) 2 W n u n x ˆ 2 + 2 α n γ f ( x n ) A x ˆ , x n + 1 x ˆ ( 1 α n γ ¯ ) 2 u n x ˆ 2 + 2 α n γ f ( x n ) f ( x ˆ ) , x n + 1 x ˆ + 2 α n γ f ( x ˆ ) A x ˆ , x n + 1 x ˆ ( 1 α n γ ¯ ) 2 ( x n x ˆ 2 x n u n 2 ) + 2 α n γ α x n x ˆ x n + 1 x ˆ + 2 α n γ f ( x ˆ ) A x ˆ x n + 1 x ˆ = ( 1 2 α n γ ¯ + ( α n γ ¯ ) 2 ) x n x ˆ 2 ( 1 α n γ ¯ ) 2 x n u n 2 + 2 α n γ α x n x ˆ x n + 1 x ˆ + 2 α n γ f ( x ˆ ) A x ˆ x n + 1 x ˆ x n x ˆ 2 + α n γ ¯ 2 x n x ˆ 2 ( 1 α n γ ¯ ) 2 x n u n 2 + 2 α n γ α x n x ˆ x n + 1 x ˆ + 2 α n γ f ( x ˆ ) A x ˆ x n + 1 x ˆ ,

and hence

( 1 α n γ ¯ ) 2 x n u n 2 x n x ˆ 2 x n + 1 x ˆ 2 + α n γ ¯ 2 x n x ˆ 2 + 2 α n γ α x n x ˆ x n + 1 x ˆ + 2 α n γ f ( x ˆ ) A x ˆ x n + 1 x ˆ x n x n + 1 ( x n x ˆ x n + 1 x ˆ ) + α n γ ¯ 2 x n x ˆ 2 + 2 α n γ α x n x ˆ x n + 1 x ˆ + 2 α n γ f ( x ˆ ) A x ˆ x n + 1 x ˆ .

Therefore, we have x n u n 0 as n.

Step 6. Show that lim sup n γf( x )A x , x n x 0.

We can choose a subsequence { x n i } of { x n } such that

lim i γ f ( x ) A x , x n i x = lim sup n γ f ( x ) A x , x n x .

Let

A( x n i )= { x H : lim sup i x n i x = inf y H lim sup i x n i y }

be the asymptotic center of { x n i }. Since { x n i } is bounded and H is a Hilbert space, it is well known that A( x n i ) is a singleton; say A( x n i )={ x ˜ }. Set

L= sup i N γ f ( x n i ) A W n i u n i

and for every xH define

Wx= lim i W n i x
(3.7)

and

T r (x)= { z H : F ( z , y ) + 1 r y z , z x 0 , y H } .

Note that

x n i W x ˜ x n i + 1 x n i + x n i + 1 W x ˜ = x n i + 1 x n i + P C [ α n i γ f ( x n i ) + ( I α n i A ) W n i u n i ] W x ˜ x n i + 1 x n i + α n i γ f ( x n i ) + ( I α n i A ) W n i u n i W x ˜ = x n i + 1 x n i + W n i u n i W x ˜ + α n i ( γ f ( x n i ) A W n i u n i ) x n i + 1 x n i + W n i u n i W n i x n i + W n i x n i W n i x ˜ + W n i x ˜ W x ˜ + α n i L x n i + 1 x n i + u n i x n i + x n i x ˜ + W n i x ˜ W x ˜ + α n i L .

By Steps 1-5, condition (C1) and (3.7), we derive

lim sup i x n i W x ˜ lim sup i u n i x n i + x n i x ˜ + W n i x ˜ W x ˜ lim sup i x n i x ˜ .

That is, W x ˜ A( x n i ). Therefore W x ˜ = x ˜ . Next, we show that x ˜ = T r x ˜ .

Note that for any xH and a,b>0, we have

F( T a x,y)+ 1 a y T a x, T a xx0,yH

and

F( T b x,y)+ 1 b y T b x, T b xx0,yH,

then

F( T a x, T b x)+ 1 a T b x T a x, T a xx0

and

F( T b x, T a x)+ 1 b T a x T b x, T a xx0.

Summing up the last inequalities and using (A2), we obtain

T a x T b x , T b x x b T a x x a 0.

Hence we have

0 T a x T b x , T b x x b a ( T a x x ) = T a x T b x , T b x T a x + T a x x b a ( T a x x ) = T a x T b x , ( T b x T a x ) + ( 1 b a ) ( T a x x ) T a x T b x 2 + | 1 b a | T a x T b x ( T a x + x ) .

We derive then

T a x T b x | b a | a ( T a x + x ) .

It follows that

x n i T r x ˜ x n i + 1 x n i + x n i + 1 T r x ˜ = x n i + 1 x n i + P C [ α n i γ f ( x n i ) + ( I α n i A ) W n i u n i ] T r x ˜ x n i + 1 x n i + α n i γ f ( x n i ) + ( I α n i A ) W n i u n i T r x ˜ = x n i + 1 x n i + W n i u n i T r x ˜ + α n i ( γ f ( x n i ) A W n i u n i ) x n i + 1 x n i + W n i u n i x n i + T r n i x n i T r n i x ˜ + x n i T r n i x n i + T r n i x ˜ T r x ˜ + α n i L x n i + 1 x n i + u n i x n i + x n i x ˜ + x n i u n i + T r n i x ˜ T r x ˜ + α n i L x n i + 1 x n i + u n i x n i + x n i x ˜ + x n i u n i + | r n i r | r ( T r x ˜ + x ˜ ) + α n i L .

By Steps 2-5, conditions (C1) and (C3), we obtain

lim sup i x n i T r x ˜ lim sup i x n i x ˜

and x ˜ = T r x ˜ . Thus x ˜ F(W)F( T r )=Ω by Lemma 2.3 and 2.9. Fix t(0,1), xH and set y= x ˜ +tx. Then

x n i x ˜ t x 2 x n i x ˜ 2 +2tx, x ˜ +tx x n i .

By the minimizing property of x ˜ and since 2 is continuous and increasing in [0,), we have

lim sup i x n i x ˜ 2 lim sup i x n i x ˜ t x 2 lim sup i x n i x ˜ 2 + 2 t lim sup i x , x ˜ + t x x n i .

Thus,

lim sup i x, x ˜ +tx x n i 0.

On the other hand,

x, x ˜ x n i =x, x ˜ +tx x n i t x 2 .

Hence we obtain

lim sup i x, x ˜ x n i = lim t 0 ( lim sup i x , x ˜ + t x x n i t x 2 ) 0.

Set x=γf( x )A x . Since x ˜ Ω, we obtain

0 lim sup i γ f ( x ) A x , x ˜ x n i γ f ( x ) A x , x ˜ x + lim i γ f ( x ) A x , x x n i lim i γ f ( x ) A x , x x n i .

So that

lim sup n γ f ( x ) A x , x n x = lim i γ f ( x ) A x , x n i x 0.

Step 7. Show that both { x n } and { u n } strongly converge to x Ω, which is the unique solution of the variational inequality (3.2). Indeed, we note that

x n + 1 x 2 = x n + 1 d n , x n + 1 x + d n x , x n + 1 x .

Since x n + 1 d n , x n + 1 x 0, we get

x n + 1 x 2 d n x , x n + 1 x = α n γ f ( x n ) + ( I α n A ) W n u n x , x n + 1 x = α n γ f ( x n ) α n γ f ( x ) + W n u n α n A W n u n x + α n A x + α n γ f ( x ) α n A x , x n + 1 x = α n γ ( f ( x n ) f ( x ) ) + ( I α n A ) ( W n u n x ) , x n + 1 x + α n γ f ( x ) A x , x n + 1 x ( α n γ f ( x n ) f ( x ) + I α n A W n u n x ) x n + 1 x + α n γ f ( x ) A x , x n + 1 x ( 1 α n ( γ ¯ γ α ) ) x n x x n + 1 x + α n γ f ( x ) A x , x n + 1 x ( [ 1 α n ( γ ¯ γ α ) ] 2 2 ) x n x 2 + 1 2 x n + 1 x 2 + α n γ f ( x ) A x , x n + 1 x .

It then follows that

x n + 1 x 2 [ 1 α n ( γ ¯ γ α ) ] x n x 2 +2 α n γ f ( x ) A x , x n + 1 x .
(3.8)

Let a n = x n x 2 , γ n = α n ( γ ¯ γα) and δ n =2 α n γf( x )A x , x n + 1 x .

Then, we can write the last inequality as

a n + 1 (1 γ n ) a n + δ n .

Note that in virtue of condition (C2), n = 1 γ n =. Moreover,

lim sup n δ n γ n = 1 γ ¯ γ α lim sup n 2 γ f ( x ) A x , x n + 1 x .

By Step 5, we obtain

lim sup n δ n γ n 0.
(3.9)

Now, applying Lemma 2.6 to (3.8), we conclude that x n x as n. Furthermore, since u n x = T r n x n T r n x x n x , we then have that u n x as n. The proof is now complete. □

Setting AI and γ=1 in Theorem 3.1, we have the following result.

Corollary 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let F:C×CR be an equilibrium bifunction satisfying the conditions:

  1. (1)

    F is monotone, that is, F(x,y)+F(y,x)0 for all x,yC;

  2. (2)

    for each x,y,zC, lim t 0 F(tz+(1t)x,y)F(x,y);

  3. (3)

    for each xC, yF(x,y) is convex and lower semicontinuous.

Let { T i } i = 1 be an infinite family of nonexpansive mappings of C into C such that i = 1 F( T i )EP(F). Suppose { α n }(0,1) and { r n }(0,) satisfy the following conditions:

  1. (1)
    lim n α n =0

    and n = 1 α n =;

  2. (2)
    lim inf n r n >0

    and lim n ( r n + 1 r n )=0.

Let f be a contraction of C into itself, and let x 0 H be given arbitrarily. Then the sequences { x n } and { y n } generated iteratively by

{ F ( y n , x ) + 1 r n x y n , y n x n 0 , x C , x n + 1 = α n f ( x n ) + ( 1 α n ) W n y n ,

converge strongly to x i = 1 F( T i )EP(F), the unique solution of the minimization problem

min x i = 1 F ( T i ) E P ( F ) 1 2 x 2 h(x),

where h is a potential function for f.

Setting F=0 in Theorem 3.1, we have the following result.

Corollary 3.3 ([11])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let { T n } n = 1 be a sequence of nonexpansive mappings from C to C such that the common fixed point set Ω= n = 1 F( T n ). Let f:CH be an α-contraction and A:HH be a strongly positive bounded linear operator with a coefficient γ ¯ >0. Let γ be a constant such that 0<γα< γ ¯ . For an arbitrary initial point x 0 belonging to C, one defines a sequence { x n } n 0 iteratively

x n + 1 = P C [ α n γ f ( x n ) + ( I α n A ) W n x n ] ,n0,
(3.10)

where { α n } is a real sequence in [0,1]. Assume that the sequence { α n } satisfies the following conditions:

(C1) lim n α n =0;

(C2) n = 1 α n =.

Then the sequence { x n } generated by (3.10) converges in norm to the unique solution x , which solves the following variational inequality:

x Ω such that ( A γ f ) x , x x ˆ 0, x ˆ Ω.
(3.11)

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Acknowledgements

The author would like to thank Asst. Prof. Dr. Rabian Wangkeeree for his useful suggestions and the referees for their valuable comments and suggestions.

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Correspondence to Kiattisak Rattanaseeha.

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Keywords

  • equilibrium problem
  • fixed point
  • nonexpansive mapping
  • variational inequality
  • strongly positive operator
  • Hilbert spaces